$\sec ^{-1}\left(\frac{x}{a}\right)-\sec ^{-1}\left(\frac{x}{b}\right)=\sec ^{-1} b-\sec ^{-1} a$
$\cos ^{-1}\left(\frac{a}{x}\right)-\cos ^{-1}\left(\frac{b}{x}\right)=\cos ^{-1}\left(\frac{1}{b}\right)-\cos ^{-1}\left(\frac{1}{a}\right)$
$\cos ^{-1}\left\{\frac{a b}{x^2}+\sqrt{1-\frac{a^2}{x^2}} \cdot \sqrt{1-\frac{b^2}{x^2}}\right\}$
$=\cos ^{-1}\left\{\frac{1}{a b}+\sqrt{1-\frac{1}{b^2}} \sqrt{1-\frac{1}{a^2}}\right\}$ $\left[\because \cos ^{-1} x-\cos ^{-1} y=\cos ^{-1}\left\{x y+\sqrt{1-x^2} \sqrt{1-y^2}\right]\right.$
$\begin{aligned} & \frac{a b}{x^2}+\frac{\sqrt{\left(x^2-a^2\right)\left(x^2-b^2\right)}}{x^2} \\ = & \frac{1}{a b}+\frac{\sqrt{\left(b^2-1\right)\left(a^2-1\right)}}{a b}\end{aligned}$
$\Rightarrow \quad a^2 b^2+a b \sqrt{\left(x^2-a^2\right)\left(x^2-b^2\right)}$
$=x^2+x^2 \sqrt{\left(a^2-1\right)\left(b^2-1\right)}$
$a b \sqrt{\left(x^2-a^2\right)\left(x^2-b^2\right)}-x^2 \sqrt{\left(a^2-1\right)}$
$\sqrt{\left(b^2-1\right)}=x^2-a^2 b^2$
in RHS $x^2-a^2 b^2$ has the minimum value zero because $x^2$ is always positive and $a^2 b^2$ is positive.
If $x^2-a^2 b^2=0$
$\begin{aligned} x^2 & =a^2 b^2 \\ x & =a b\end{aligned}$